Question

1. Чему равна масса сульфида натрия, полученного при взаимодействии 6,9 г натрия с серой? Условие + решение,спасибо

Answer 1

Ответ: решение во вложении.

Объяснение:

Answer 2

I can help you with your chemistry question.

To find the mass of sodium sulfide, we need to write the balanced chemical equation for the reaction of sodium and sulfur:

$$2Na + S \rightarrow Na_2S$$

According to the equation, 2 moles of sodium react with 1 mole of sulfur to produce 1 mole of sodium sulfide. We can use the molar masses of the substances to convert grams to moles:

$$M(Na) = 23 g/mol$$ $$M(S) = 32 g/mol$$ $$M(Na_2S) = 78 g/mol$$

$$n(Na) = \frac{6.9 g}{23 g/mol} = 0.3 mol$$ $$n(S) = \frac{m(S)}{32 g/mol}$$ $$n(Na_2S) = \frac{m(Na_2S)}{78 g/mol}$$

Since sodium is the limiting reactant, we can use its amount to calculate the amount of sodium sulfide:

$$n(Na_2S) = \frac{n(Na)}{2} = 0.15 mol$$

Then, we can convert moles to grams:

$$m(Na_2S) = n(Na_2S) \times M(Na_2S) = 0.15 mol \times 78 g/mol = 11.7 g$$

Therefore, the mass of sodium sulfide is 11.7 grams.

I hope this helps. If you want to learn more about chemical reactions, you can check out this [website](https://infourok.ru/samostoyatelnaya-rabota-po-himii-dlya-uchaschihsya-klassa-po-teme-reshenie-raschetnih-zadach-po-uravneniyu-reakcii-samostoyateln-3646059.html).